To a solution containing 10.4 g of barium chloride was added a solution containing 9.8 g of sulfuric acid.

To a solution containing 10.4 g of barium chloride was added a solution containing 9.8 g of sulfuric acid. The precipitate was filtered off and dried. how much dry sediment did you get? what substances will be in solution

Barium chloride reacts with sulfuric acid. This forms a water-insoluble barium sulfate salt, which precipitates. The reaction is described by the following chemical reaction equation.

BaCl2 + H2SO4 = BaSO4 + 2HCl;

Barium chloride reacts with sulfuric acid in equal molar amounts. In this case, the same amount of insoluble barium sulfate is synthesized.

Let’s find the chemical amount of sulfuric acid.

M H2SO4 = 2 + 32 + 16 x 4 = 98 grams / mol; N H2SO4 = 9.8 / 98 = 0.1 mol;

Find the chemical amount of barium chloride.

M BaCl2 = 137 + 35.5 x 2 = 208 grams / mol; N BaCl2 = 10.4 / 208 = 0.05 mol;

The same amount of barium sulfate will be synthesized (sulfuric acid is taken in excess)

Let’s calculate its weight.

M BaSO4 = 137 + 32 + 16 x 4 = 233 grams / mol; m BaSO4 = 0.05 x 233 = 11.65 grams;

Barium sulfate will precipitate. Synthesized hydrochloric acid and excess sulfuric acid will remain in the solution.